Question: Is ${534178}$ divisible by $9$ ?
Explanation: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {534178}= &&{5}\cdot100000+ \\&&{3}\cdot10000+ \\&&{4}\cdot1000+ \\&&{1}\cdot100+ \\&&{7}\cdot10+ \\&&{8}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {534178}= &&{5}(99999+1)+ \\&&{3}(9999+1)+ \\&&{4}(999+1)+ \\&&{1}(99+1)+ \\&&{7}(9+1)+ \\&&{8} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {534178}= &&\gray{5\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {5}+{3}+{4}+{1}+{7}+{8} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${534178}$ is divisible by $9$ if ${ 5}+{3}+{4}+{1}+{7}+{8}$ is divisible by $9$ Add the digits of ${534178}$ $ {5}+{3}+{4}+{1}+{7}+{8} = {28} $ If ${28}$ is divisible by $9$ , then ${534178}$ must also be divisible by $9$ ${28}$ is not divisible by $9$, therefore ${534178}$ must not be divisible by $9$.